Solving differential equation

  • Thread starter wormhole
  • Start date
  • #1
29
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i'm trying to find a mirror shape which focuses a light at some specific point [itex]x_0[/itex]
the initial equation i derived for determining the shape of the mirror is:
(assuming that light rays fall parallel to x axis - light source is very far from the mirror)
f(x) is the shape i'm trying to determine

[tex]
x_0=-\frac{f(x)-\tan(2\arctan(\frac{df}{dx}))x}{\tan(2\arctan(\frac{df}{dx}))}
[/tex]

basicly this is an expression for a line passing through point [itex]x_0[/itex] and point on
f(x) where light reflected.
so [itex]\tan(2\arctan(\frac{df}{dx}))[/itex] is a incline of this line


from the initial equation i got to this point and i'm not sure what to do next::confused:

[tex]
\frac{f(x)}{x-x_0}=\tan(2\arctan(\frac{df}{dx}))
[/tex]
 
Last edited:

Answers and Replies

  • #2
29
0
ok, i took the formula
[tex]
\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}
[/tex]

using this formula i can simplify the equation to become:

[tex]
\frac{2\frac{df}{dx}}{1-{(\frac{df}{dx})}^2}=\frac{f(x)}{x-x_0}
[/tex]

is it possible to solve it?
 
  • #3
185
4
i get [itex] f = \pm \sqrt{ (x+c)^2 - (x-x_0)^2 } [/itex] as
a solution. (where c is an arbitrary constant, hopefully
not the same as -x0.)
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,847
966
qbert said:
i get [itex] f = \pm \sqrt{ (x+c)^2 - (x-x_0)^2 } [/itex] as
a solution. (where c is an arbitrary constant, hopefully
not the same as -x0.)

Well, I'm impressed with what you have done so far!
I assume you mean the light is focussed at the point (x0,0).

You might as well multiply out the terms in the square root:
[tex]f(x)= \sqrt{x^2+ 2cx+ c^2- x^2+2x_0x- x_0^2}[/tex]

You can't determine c from the information given: any parabola with focus at (x0,0) will focus light as required.
And don't write [itex]f(x)= \pm \sqrt{...}[/itex]. A function is single valued. You can, of course, write x as a function of y.
 

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